POJ 1066 Treasure Hunt【线段相交】

思路：枚举四边墙的门的中点，与终点连成一条线段，判断与其相交的线段的个数。最小的加一即为答案。

我是傻逼，一个数组越界调了两个小时。

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
using namespace std;
const int N=;
const double eps=1e-;
int sgn(double x){
    if(fabs(x)<eps)    return ;
    if(x>)    return ;
    return -;
}
struct point{
    double x,y;
    point(){}
    point(double x_,double y_){
        x=x_,y=y_;
    }
    point operator -(const point &b)const{
        return point(x-b.x,y-b.y);
    }
    double operator *(const point &b)const{
        return x*b.x+y*b.y;
    }
    double operator ^(const point &b)const{
        return x*b.y-y*b.x;
    }
}po[N];
struct line{
    point s,e;
    line(){}
    line(point s_,point e_){
        s=s_,e=e_;
    }
}li[N];
double cal(point p0,point p1,point p2){//叉积
    return (p1-p0)^(p2-p0);
}
int xj(line a,line b){//判断两线段是否相交
    point A=a.s,B=a.e,C=b.s,D=b.e;
    return
    max(A.x,B.x)>=min(C.x,D.x) &&
    max(C.x,D.x)>=min(A.x,B.x) &&
    max(A.y,B.y)>=min(C.y,D.y) &&
    max(C.y,D.y)>=min(A.y,B.y) &&
    sgn(cal(C,A,B))*sgn(cal(D,A,B))<= &&
    sgn(cal(A,C,D))*sgn(cal(B,C,D))<=;
}
int p[][];
int main(){
    int n,i,j,js;
    double x1,y1,x2,y2;
    while(~scanf("%d",&n)){
        memset(p,,sizeof(p));js=;
        for(i=;i<=n;i++){
            scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
            p[(int)x1][(int)y1]=p[(int)x2][(int)y2]=;
            li[i]=line(point(x1,y1),point(x2,y2));
        }
        scanf("%lf%lf",&x1,&y1);
        point P=point(x1,y1);
        p[][]=;
        p[][]=;
        p[][]=;
        p[][]=;
        int x0,y0;
        for(y0=,i=;i<=;i++){
            if(p[][i]){
                po[++js]=point(,(i+y0)/2.0);
                y0=i;
            }
        }
        for(x0=,i=;i<=;i++){
            if(p[i][]){
                po[++js]=point((i+x0)/2.0,);
                x0=i;
            }
        }
        for(y0=,i=;i<=;i++){
            if(p[][i]){
                po[++js]=point(,(i+y0)/2.0);
                y0=i;
            }
        }
        for(x0=,i=;i<=;i++){
            if(p[i][]){
                po[++js]=point((i+x0)/2.0,);
                x0=i;
            }
        }
        int ans=;
        for(i=;i<=js;i++){
            line L=line(P,po[i]);
            int cnt=;
            for(j=;j<=n;j++){
                if(xj(L,li[j])){
                    cnt++;
                }
            }
            if(cnt<ans)    ans=cnt;
        }
        printf("Number of doors = %d\n",ans+);
    }
    return ;
}